EZ logarithm V2

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fruitmonster97

2491 posts

fruitmonster97

#1 h Feb 2, 2024, 5:27 PM • 1 Y

Y by MathFan335

Real numbers $x$ and $y$ with $x,y>1$ satisfy $\log_x(y^x)=\log_y(x^{4y})=10.$ What is the value of $xy$ ?

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Bluesoul

894 posts

Bluesoul

#2 h Feb 2, 2024, 5:27 PM • 1 Y

Y by sharknavy75

4xy=100, xy=25

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balllightning37

389 posts

balllightning37

#3 h Feb 2, 2024, 5:27 PM • 1 Y

Y by Neeka_s08

I got 025. Anyone confirm?

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mitsuihisashi14

121 posts

mitsuihisashi14

#4 h Feb 2, 2024, 5:28 PM

Y by

Anyone confirm 025?

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gladIasked

648 posts

gladIasked

#5 h Feb 2, 2024, 5:28 PM • 1 Y

Y by isache

025, anyone? you could just solve for xy directly.

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EvanZ

187 posts

EvanZ

#6 h Feb 2, 2024, 5:28 PM • 2 Y

Y by fura3334, dbnl

i trolled this for 2 hours and ended up guessing 25...

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megarnie

5606 posts

megarnie

#8 h Feb 2, 2024, 5:30 PM

Y by

Let $y = x^k$ . We want to find $x^{k+1}$ .

We have $10 = \log_x(y^x) = xk$ and $10 =\log_y(x^{4y} ) = \frac{4y}{k} = \frac{4x^k}{k}$ .

Multiplying equations gives $4x^{k+1} = 10 \cdot 10$ , so answer is $\boxed{025}$ .

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cinnamon_e

703 posts

cinnamon_e

#9 h Feb 2, 2024, 5:41 PM • 5 Y

Y by MathFan335, Jack_w, dbnl, lucifer9838, Math_DM

this has to be the easiest aime problem ever

just multiply to get $x\log_xy\cdot 4y\log_yx=4xy=100\implies xy=\boxed{025}$ .

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shendrew7

795 posts

shendrew7

#10 h Feb 2, 2024, 5:47 PM

Y by

Our answer is
$xy = \frac 14 \cdot 4xy = \frac 14 (x \log_xy)(4y \log_yx) = = \frac 14 \cdot 10 \cdot 10 = \boxed{025}. \quad \blacksquare$

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williamxiao

2514 posts

williamxiao

#11 h Feb 2, 2024, 6:03 PM

Y by

By log rules, we have $x\log_{x}y = 4y\log_{y}x = 10$ . Letting $\log_{x}y = z$ , we have $xz = \frac{4y}{z}= 10$ . We get $z = \frac{10}{x} = \frac{4y}{10}$ , so $4xy=100$ and $xy=025$

This post has been edited 1 time. Last edited by williamxiao, Feb 2, 2024, 6:04 PM

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bluelinfish

1449 posts

bluelinfish

#12 h Feb 2, 2024, 6:11 PM

Y by

Simplify to get that $x\log_x(y) = 4y\log_y(x) = 10$ . Therefore $100 = 4xy \log_x(y)\log_y(x) = 4xy$ , implying that $xy = \boxed{025}$ .

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cappucher

94 posts

cappucher

#13 h Feb 2, 2024, 6:14 PM

Y by

I was worried on how easy this was, but $025$ is right...

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MC413551

2228 posts

MC413551

#14 h Feb 2, 2024, 6:29 PM • 1 Y

Y by matchaisgreen

x^10=y^x
y^10=x^4y
take 10th root of first equatoin to get
x=y^(x/10)
substitute into second equation
y^10=y^4xy/10
10=4xy/10
100=4xy
xy=25

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gicyuraok2

1059 posts

gicyuraok2

#15 h Feb 2, 2024, 7:19 PM

Y by

bro last year i guessed 25 on all my unanswered questions, and i was going to do the same thing this year until i solved this problem first. wut

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exp-ipi-1

1074 posts

exp-ipi-1

#16 h Feb 2, 2024, 7:22 PM

Y by

MC413551 wrote:

x^10=y^x
y^10=x^4y
take 10th root of first equatoin to get
x=y^(x/10)
substitute into second equation
y^10=y^4xy/10
10=4xy/10
100=4xy
xy=25

yeah thats what i did

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brainfertilzer

1831 posts

brainfertilzer

#17 h Feb 2, 2024, 8:09 PM

Y by

??? $x\log_x y =10, 4y\log_y x = 10\implies 4xy(\log_xy)(\log_y x) = 100\implies 4xy = 100\implies xy = \boxed{025}$

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plang2008

337 posts

plang2008

#18 h Feb 2, 2024, 8:16 PM

Y by

I can't be the only one who got the answer without even meaning to right?

Rewrite without logs as $x^{10} = y^x$ and $y^{10} = x^{4y}$ . Raise the second equation to the $2.5$ th power to get $y^{25} = x^{10y} = y^{xy}$ so $xy = \boxed{25}$ .

After I got $xy = 25$ I attempted to find $x$ and $y$ before realizing what it was asking for...

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Mebookreader

1103 posts

Mebookreader

#19 h Feb 2, 2024, 9:54 PM

Y by

Bruh I didn't know how to do this.

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zhu1223

2 posts

zhu1223

#20 h Feb 2, 2024, 10:07 PM

Y by

I agree this should be 25

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Hayabusa1

478 posts

Hayabusa1

#21 h Feb 2, 2024, 10:08 PM

Y by

$\log_x y=\frac{10}{x}\implies 4y\log_y x=4y\cdot \frac{x}{10}=10$ . Thus, $xy=\frac{10\cdot 10}{4}=\boxed{025}$ .

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jasperE3

11293 posts

jasperE3

#22 h Feb 2, 2024, 11:16 PM • 2 Y

Y by FelipeGigena, MrDanny

fruitmonster97 wrote:

Real numbers $x$ and $y$ with $x,y>1$ satisfy $\log_x(y^x)=\log_y(x^{4y})=10.$ What is the value of $xy$ ?

Using $\log$ rules:
$x\log_xy=4y\log_yx=10.$ Noticing the $\log_xy$ and $\log_yx$ , recalling the identity $\log_xy\cdot\log_yx=1$ we multiply $x\log_xy$ and $4y\log_yx$ to get:
$4xy=x\log_xy\cdot4y\log_yx=10\cdot10=100.$ Then $xy=\boxed{025}$ .

This post has been edited 1 time. Last edited by jasperE3, Feb 3, 2024, 12:37 AM

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Arrowhead575

2281 posts

Arrowhead575

#23 h Feb 3, 2024, 12:00 AM

Y by

Most straightforward AIME question of all time?

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exp-ipi-1

1074 posts

exp-ipi-1

#24 h Feb 3, 2024, 12:02 AM

Y by

Arrowhead575 wrote:

Most straightforward AIME question of all time?

nah 2023 AIME 2 P1

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mathboy282

2989 posts

mathboy282

#25 h Feb 3, 2024, 12:03 AM

Y by

fruitmonster97 wrote:

Real numbers $x$ and $y$ with $x,y>1$ satisfy $\log_x(y^x)=\log_y(x^{4y})=10.$ What is the value of $xy$ ?

By Change of Base notice that log_x(y)=1/log_y(x). Thus, we have xlog_x(y)=(4y)log_y(x)=10. Then we get xlog_x(y)=4y/log_x(y)=10. Solving for log_x(y) we get log_x(y)=10/x. Then: 4y/(10/x)=10 => 2xy/5=10 => xy=25.

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orangesyrup

130 posts

orangesyrup

#26 h Feb 3, 2024, 3:09 PM

Y by

the ans is 025

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xHypotenuse

778 posts

xHypotenuse

#27 h Feb 4, 2024, 4:19 AM

Y by

this shouldve been problem 1

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am07

316 posts

am07

#28 h Feb 4, 2024, 2:47 PM

Y by

Easiest problem on the test

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AshAuktober

1004 posts

AshAuktober

#29 h Feb 6, 2024, 11:59 AM

Y by

When in doubt, spam log properties.
Observe that $\log_x(y^x) = x \frac{\log y}{\log x} = 10, \log_y(x^{4y}) = 4y\frac{\log x}{\log y} = 10$ . Multiplying these two, we get $4xy = 100 \implies xy = \boxed{025}$ .

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sharknavy75

700 posts

sharknavy75

#30 h Feb 6, 2024, 1:43 PM

Y by

I did a very stupid solution.

I did the following
x^10 = y^x
y^10 = x^4y
multiply the 2 equations and get
x^10*y^10 = y^x*x^4y
1 = y^(x-10)x^(4y-10)
x-10 = 0 and 4y-10=0
x = 10 and y = 2.5
xy = 25

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megahertz13

3183 posts

megahertz13

#32 h Feb 24, 2024, 9:47 PM

Y by

Video Solution & More

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Soumya_cena

17 posts

Soumya_cena

#33 h Mar 10, 2024, 4:12 AM

Y by

fruitmonster97 wrote:

Real numbers $x$ and $y$ with $x,y>1$ satisfy $\log_x(y^x)=\log_y(x^{4y})=10.$ What is the value of $xy$ ?

Quite easy with logarithmic properties

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Mr.Sharkman

498 posts

Mr.Sharkman

#34 h Apr 2, 2024, 2:25 AM • 1 Y

Y by blueprimes

Solution

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aidan0626

1897 posts

aidan0626

#35 h Apr 2, 2024, 4:45 AM • 1 Y

Y by Math_DM

pov: you get stuck on this problem in test because you didn't notice the "=10" part until like 10 minutes after
don't remember how i solved this in contest, i didn't notice you can just multiply them lol

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A04572

9 posts

A04572

#36 h Apr 21, 2024, 12:40 PM

Y by

xlogx y=4ylogyx=10 logxy*logyx=1, 4xy=100 xy=25 Can we solve for x and y directly?

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ProMaskedVictor

43 posts

ProMaskedVictor

#37 h Jun 11, 2024, 1:27 PM

Y by

$\log_x (y^x) =x\log_x y=10 -(i)$
$\log_y (x^{4y}) =4y\log_y x=10 -(ii)$
So, $(i) \times (ii):$
$4xy=100 \implies \boxed{xy=25}$

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Rio_24

2 posts

Rio_24

#38 h Jun 29, 2024, 3:53 AM

Y by

The Answer xy = 25

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meduh6849

354 posts

meduh6849

#39 h Jul 9, 2024, 12:00 AM

Y by

Yet another 025... once again, reinforcing its status as the most common AIME answer.
https://artofproblemsolving.com/community/c5h2018116

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gracemoon124

872 posts

gracemoon124

#40 h Aug 16, 2024, 4:42 AM

Y by

storage

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PYM7

5 posts

PYM7

#41 h Dec 11, 2024, 5:02 AM

Y by

Let $y = x^{k}$ , where $k$ is a non-negative real number. The condition of the problem is equivalent to $xk = \frac{4x^{k}}{k} = 10$ . If we multiply $xk$ by all the terms here, we get $(xk)^2 = 4x^{k+1}$ . Since $xy = x^{k+1}$ , $xy = \frac{(xk)^2}{4} = \frac{100}{4} = 25$